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definite integral

paulmdrdo

Active member
May 13, 2013
386
i tried to solve this definite integral but i keep on getting an invalid answer. please check my error.

$\displaystyle \int_{-3}^{-2}\frac{y+2}{y^2+4y}dy$

$\displaystyle u=y^2+4y$
$\displaystyle du=2y+4dy$
$\displaystyle dy=\frac{du}{2y+4}$

$\displaystyle \frac{1}{2}\int\frac{y+2}{u}\times \frac{du}{2(y+2)}=\frac{1}{2}\int\frac{du}{u}= \frac{1}{2}\ln|u|+c= \frac{1}{2}\ln|y^2+4y|+c$

when i calculate the definite integral i always get an error.

$\displaystyle\frac{1}{2}\ln|(-2)^2+4(-2)|-\frac{1}{2}\ln|(-3)^2+4(-3)| = ?$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What you have done so far is correct...what is your final answer?
 

paulmdrdo

Active member
May 13, 2013
386
$\displaystyle\frac{1}{2}\ln|(-2)^2+4(-2)|-\frac{1}{2}\ln|(-3)^2+4(-3)| = \frac{1}{2}\ln|-4|-\frac{1}{2}\ln|-3|= ?$ i punch this in the calculator it gives me an error.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Use:

\(\displaystyle |-4|=4,\,|-3|=3\)

to write your solution.